A man is known to speak truth$3$ out of$4$ times. He takes out a card at random from a well shuffled pack of$52$ playing cards, and reports it is  a  king. The probability that its actually a king is

$E1 :$card drawn is a king

 $E2 :$ card drawn is not a king

$A :$ man reports it is a king.

We have\

$P\left(E_1\right)=\frac{4}{52}=\frac{1}{13},P\left(E_2\right)=\frac{12}{13}$

$P\left(A\mid E_1\right)=\frac{3}{4}$ and $P\left(A\mid E_2\right)=\frac{1}{4}$

By the Bayes’ rule

$\begin{array}{l}P\left(E_1\mid A\right)=\frac{P\left(E_1\right)P\left(A\mid E_1\right)}{P\left(E_1\right)P\left(A\mid E_1\right)+P\left(E_2\right)P\left(A\mid E_2\right)}\\ =\frac{\left(\frac{1}{13}\right)\left(\frac{3}{4}\right)}{\left(\frac{1}{13}\right)\left(\frac{3}{4}\right)+\left(\frac{12}{13}\right)\left(\frac{1}{4}\right)}=\frac{1}{5}\end{array}$